solve-for-x-and-y-tan-2-pi-x-y-cot-2-pi-x-y-1-2x-x-2-1-

Question Number 90739 by ajfour last updated on 25/Apr/20

s o l v e f o r x a n d y

\tan^{2} [π (x + y)] + \cot^{2} [π (x + y)]

= 1 + \sqrt{\frac{2 x}{x^{2} + 1}}

Commented by me2love2math last updated on 26/Apr/20

m a x v a l u e o r m i n i m u m ?

Commented by ajfour last updated on 26/Apr/20

g e n e r a l v a l u e s .

Answered by mr W last updated on 26/Apr/20

\tan^{2} [π (x + y)] + \cot^{2} [π (x + y)] = 1 + \sqrt{\frac{2 x}{x^{2} + 1}}

\tan^{2} [π (x + y)] + \cot^{2} [π (x + y)] - 2 = \sqrt{\frac{2 x}{x^{2} + 1}} - 1

{\tan [π (x + y)] - \cot [π (x + y)]}^{2} = \sqrt{\frac{2 x}{x^{2} + 1}} - 1

L H S = {\dots}^{2} ⩾ 0

R H S = \sqrt{\frac{2 x}{x^{2} + 1}} - 1 ⩽ 0

\Rightarrow L H S = R H S = 0

\sqrt{\frac{2 x}{x^{2} + 1}} - 1 = 0

\Rightarrow x = 1

\tan [π (x + y)] - \cot [π (x + y)] = 0

\tan^{2} [π (x + y)] = 1

\tan [π (x + y)] = \pm 1

\Rightarrow π (x + y) = k π \pm \frac{π}{4}

\Rightarrow x + y = k \pm \frac{1}{4}

\Rightarrow y = k - 1 \pm \frac{1}{4}

\Rightarrow y = n \pm \frac{1}{4} w i t h n = a n y i n t e g e r

o r

\Rightarrow y = \frac{n}{2} + \frac{1}{4} w i t h n = a n y i n t e g e r

Commented by mr W last updated on 26/Apr/20

a t r i c k y q u e s t i o n! i {d i d n}^{'} t k n o w h o w

t o s o l v e s t r a i g h t a w a y .

Commented by ajfour last updated on 26/Apr/20

S i r a n s w e r i s x = 1, y = \frac{2 n - 3}{4} .

Commented by mr W last updated on 26/Apr/20

y = n \pm \frac{1}{4} a n d y = \frac{2 n - 3}{4} a r e i d e n t i c a l .

Commented by ajfour last updated on 26/Apr/20

i d i d n o t d o u b t, b u t {c o u l d n}^{'} t

u n d e r s t a n d v e r y w e l l, s i r .

A n y w a y t h a n k s, g r e a t s o l v i n g!

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